The answer is: " 56 g CaCl₂ " .
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Explanation:
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2.0 M CaCl₂ = 2.0 mol CaCl₂ / L ;
Since: "M" = "Molarity" (measurement of concentration);
= moles of solute per L {"Liter"} of solution.
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Note the exact conversion: 1000 mL = 1 L .
Given: 250 mL ;
250 mL = ? L ? ;
250 mL * (1 L / 1000 L) = (250/1000) L = 0.25 L .
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(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol = 0.50 mol CaCl₂ ;
We have: 0.50 mol CaCl₂ ; Convert to "g" (grams):
→ 0.50 mol CaCl₂ .
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1 mol CaCl₂ = ? g ?
From the Periodic Table of Elements:
1 mol Ca = 40.08 g
1 mol Cl = <span>35.45 g .
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There are 2 atoms of Cl in " CaCl₂ " ;
→ Note the subscript, "2", in the " Cl₂ " ;
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So, to calculate the molar mass of "CaCl₂" :
40.08 g + 2(35.45 g) =
40.08 g + 70.90 g = 110.98 g ; round to 4 significant figures;
→ round to 111 g/mol .
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So:
→ 0.50 mol CaCl₂ = ? g CaCl₂ ? ;
→ 0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;
= (0.50) * (111 g) CaCl₂ ;
= 55.5 g CaCl₂ ;
→ round to 2 significant figures;
→ 56 g CaCl₂ .
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The answer is: " 56 g CaCl₂ " .
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Just 2 valence electrons.
Hydrogen already has one to start with, as well. With the exception of hydrogen and helium, all other atoms need 8 valence e-
Answer:
d) carbon dioxide and water
Answer:
1.1 mol
Explanation:
n=m/M, where n is moles, m is mass, and M is molar mass.
M of CO2 = 12.01+16.00+16.00 = 44.01g/mol
n=50g/44.01g/mol
n = 1.13610543 mol
n ≈ 1.1 mol
Hope that helps
Answer:
The volume of the gas is 2.80 L.
Explanation:
An ideal gas is a theoretical gas that is considered to be made up of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The Pressure (P) of a gas on the walls of the container that contains it, the Volume (V) it occupies, the Temperature (T) at which it is located and the amount of substance it contains (number of moles, n) are related from the equation known as Equation of State of Ideal Gases:
P*V = n*R*T
where R is the constant of ideal gases.
In this case:
- P= 2 atm
- V= ?
- n=0.223 moles
- R= 0.0821
- T=33 °C= 306 °K (being O°C= 273°K)
Replacing:
2 atm* V= 0.223 moles*0.0821 * 306 K
Solving:
V= 2.80 L
<u><em>The volume of the gas is 2.80 L.</em></u>
